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Post by liffy99 on May 6, 2018 17:34:09 GMT
I always thought Watts = Volts x Amps.
How do I calculate the voltage output of an amp ?
For example, my Lyngdorf delivers 200W into 8 ohms (375 into 4) and 40 amps peak current. I think it also quotes a max 33V output.
Now, 40x33 = a damned sight more than 200.
What's the secret ?
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Post by Clive on May 6, 2018 19:27:12 GMT
You need to use continuous current, not peak.
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Post by liffy99 on May 6, 2018 19:55:30 GMT
So what would that be ? Square root of 40?
So say 33x 6.6 = about 215
That looks a lot better ( if continuous is a sq root function).
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Post by MartinT on May 6, 2018 22:17:05 GMT
The voltage output of an amplifier is going to be close to the power supply rails to the output transistors. You could read them with a meter and subtract a little for transistor losses (a little more for output darlingtons). Alternatively, use an oscilloscope and read the output waveform just before the onset of clipping (flattening of the sinewave) using a test tone. You really need to do it with a load, which would be a simulated speaker load.
To calculate RMS power, you need the RMS voltage which is VRMS = VPeak / √2 for a sinewave.
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Deleted
Deleted Member
Posts: 0
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Post by Deleted on May 24, 2018 11:46:04 GMT
The voltage output of an amplifier is going to be close to the power supply rails to the output transistors. You could read them with a meter and subtract a little for transistor losses (a little more for output darlingtons). Alternatively, use an oscilloscope and read the output waveform just before the onset of clipping (flattening of the sinewave) using a test tone. You really need to do it with a load, which would be a simulated speaker load. To calculate RMS power, you need the RMS voltage which is V RMS = V Peak / √2 for a sinewave. Great Martin but if you use FET,s this changes and it will depend on the RdOn and also some circuit using transistors/darlingtons emitter resistors and on some FET design a source resistor. Now a few odd designs are now using IGBT, and there is the dreaded D Class amps. But it is all fun, oh here is my Yellow Cab bye. |
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Post by MartinT on May 24, 2018 12:00:24 GMT
Oh yes, all bets are off for FETs. I was only thinking of bipolar trannies in standard push/pull mode.
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Post by Firebottle on May 27, 2018 7:42:12 GMT
For a sine wave the RMS power is equivalent to the average power.
To calculate average power take the peak voltage output (half of the peak to peak), square it then divide by twice the output load resistance.
P=Vpeak(squared)/2Rload
Working backwards from 200W (assuming RMS) into 8 ohms:
Vpeak(squared) = 200x16 = 3200. So the peak output voltage is the square root, 56V.
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abcaudio
Rank: Soloist
Building cables and tea drinking
Posts: 11
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Post by abcaudio on Jun 4, 2018 10:54:08 GMT
The voltage output of an amplifier is going to be close to the power supply rails to the output transistors. You could read them with a meter and subtract a little for transistor losses (a little more for output darlingtons). Alternatively, use an oscilloscope and read the output waveform just before the onset of clipping (flattening of the sinewave) using a test tone. You really need to do it with a load, which would be a simulated speaker load. To calculate RMS power, you need the RMS voltage which is V RMS = V Peak / √2 for a sinewave. Great Martin but if you use FET,s this changes and it will depend on the RdOn and also some circuit using transistors/darlingtons emitter resistors and on some FET design a source resistor. Now a few odd designs are now using IGBT, and there is the dreaded D Class amps. But it is all fun, oh here is my Yellow Cab bye. This is all beyond me, but having built a pair of IGBT power amps with Col, I can say they seem top produce a whole lot more than 27W! I have made them clip (in a large room), but it hurt. I reckon amp manufacturers ought to take a leaf from Rolls Royce (hand)book when describing power output: "Adequate". |
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Post by MartinT on Jun 4, 2018 12:38:56 GMT
I don't care about power ratings any more.
Instantaneous current delivery is much more important for me (especially with 4 ohm speakers).
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Post by liffy99 on Jun 17, 2018 18:51:46 GMT
And, as a stat owner, voltage is more important than current to me.
56v seems just about perfect for my Quads.
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Post by MartinT on Jun 17, 2018 19:16:35 GMT
Yes, indeed!
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